Proof of n - \lceil \dfrac{n}{2} \rceil = \lfloor \dfrac{n}{2} \rfloor (Latex test - MathJax)
The key here is to divide the cases, it is not really difficult at all
To proof : n - \lceil \dfrac{n}{2} \rceil = \lfloor \dfrac{n}{2} \rfloor
Case 1. n = 2k + 1
n - \lceil \dfrac{n}{2} \rceil = 2k + 1 - ( k+1) = \lfloor k + \dfrac{1}{2} \rfloor = k
Both the LHS and RHS are k therefore, the equality stands true.
Case 2. n = 2k
n - \lceil \dfrac{n}{2} \rceil = 2k - (k) = \lfloor \dfrac{2k}{2} \rfloor = k
Both the LHS and RHS are k therefore, the equality stands true.
Thus, we have shown n - \lceil \dfrac{n}{2} \rceil = \lfloor \dfrac{n}{2} \rfloor
To proof : n - \lceil \dfrac{n}{2} \rceil = \lfloor \dfrac{n}{2} \rfloor
Case 1. n = 2k + 1
n - \lceil \dfrac{n}{2} \rceil = 2k + 1 - ( k+1) = \lfloor k + \dfrac{1}{2} \rfloor = k
Both the LHS and RHS are k therefore, the equality stands true.
Case 2. n = 2k
n - \lceil \dfrac{n}{2} \rceil = 2k - (k) = \lfloor \dfrac{2k}{2} \rfloor = k
Both the LHS and RHS are k therefore, the equality stands true.
Thus, we have shown n - \lceil \dfrac{n}{2} \rceil = \lfloor \dfrac{n}{2} \rfloor
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